Bài tập: 1. CMR: $cos(a+b).cos(a-b)=cos^2a-sin^2a$ 2. CMR: $\frac{1-cosx+cos2x}{sin2x-sinx}=cotx$ 07/12/2021 Bởi Cora Bài tập: 1. CMR: $cos(a+b).cos(a-b)=cos^2a-sin^2a$ 2. CMR: $\frac{1-cosx+cos2x}{sin2x-sinx}=cotx$
1. `cos(a+b).cos(a-b)` `=(cosa.cosb-sina.sinb).(cosa.cosb+sina.sinb)` `=cos^2a.cos^2b-sin^2a.sin^2b` `=cos^2b.(1-sin^2a)-sin^2a.(1-cos^2b)` `=cos^2b-cos^2b.sin^2a-sin^2a+cos^2b.sin^2a` `=cos^2b-sin^2a` (đpcm) 2. `\frac{1-cosx+cos2x}{sin2x-sinx}` `=\frac{1-cosx+2cos^2x-1}{2.sinx.cosx-sinx}` `=\frac{cosx(2cosx-1)}{sinx(2cosx-1)}` `=\frac{cosx}{sinx}=cotx` (đpcm) Bình luận
1.
`cos(a+b).cos(a-b)`
`=(cosa.cosb-sina.sinb).(cosa.cosb+sina.sinb)`
`=cos^2a.cos^2b-sin^2a.sin^2b`
`=cos^2b.(1-sin^2a)-sin^2a.(1-cos^2b)`
`=cos^2b-cos^2b.sin^2a-sin^2a+cos^2b.sin^2a`
`=cos^2b-sin^2a` (đpcm)
2.
`\frac{1-cosx+cos2x}{sin2x-sinx}`
`=\frac{1-cosx+2cos^2x-1}{2.sinx.cosx-sinx}`
`=\frac{cosx(2cosx-1)}{sinx(2cosx-1)}`
`=\frac{cosx}{sinx}=cotx` (đpcm)